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Periodic boundary value problems for second-order functional differential equations with impulse

Abstract

In this paper, we study the existence of multiple positive solutions of the second-order periodic boundary value problems for functional differential equations with impulse. The proof of our main results is based upon the fixed point index theorem in cones.

1 Introduction

In this paper, we will consider the existence of positive solutions for boundary value problems of second order impulsive functional differential equations of the form

{ ( ρ ( t ) u ( t ) ) + f ( t , u t ) = 0 , t J , t t k , k = 1 , , m , Δ u ( t k ) = I k ( u t k ) , k = 1 , 2 , , m , u 0 = φ , u ( T ) = A ,
(1.1)

where J=[0,T], f:J× C τ R is a continuous function, φ C τ ( C τ be given in Section 2), τ0, ρ(t)C(J,(0,)), u t C τ , u t (θ)=u(t+θ), θ[τ,0]. I k C( C τ ,R), 0= t 0 < t 1 < t 2 << t m < t m + 1 =T, J =(0,T){ t 1 ,, t m }. Δ u ( t k )= u ( t k + ) u ( t k ), u ( t k + ) ( u ( t k )) denote the right limit (left limit) of u (t) at t= t k , and AR=(,+).

Impulsive differential equations describe processes which experience a sudden change of their state at certain moments. The theory of impulse differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes arising in phenomena studied in physics, population dynamics, chemical technology and biotechnology; see [13].

Many papers have been published about the existence analysis of periodic boundary value problems of first and second order for ordinary or functional or integro-differential equations with impulsive. We refer the readers to the papers [423]. For instance, in [6], He and Yu investigated the following problem:

{ u ( t ) = f ( t , u ( t ) , u t ) , t t k , 0 < t < T , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( t ) = u ( 0 ) , t [ τ , 0 ) , u ( 0 ) = u ( T ) .

By using the coincidence degree, Dong [4] studied the following periodic boundary value problems (PBVP) for first-order functional differential equations with impulse:

{ u ( t ) = f ( t , u t ) , t t k , 0 < t < T , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( 0 ) = u ( T ) .
(1.2)

It is remarkable that the author required u(t)=u(t+1) for t[τ,0]. The author also obtained the existence of one solution of PBVP (1.2).

To study periodic boundary value problems for first and second order functional differential equations with impulse, the approaches used in [4, 68, 1316, 2023] are the monotone iterative technique and the method of upper and lower solutions. What they obtained is the existence of at least one solution if there is a pair of upper and lower solutions. However, in some cases it is difficult to find upper and lower solutions for general differential equations.

As we know, the fixed point theorem of cone expression and compression is extensively used to study the existence of multiple solutions of boundary value problems for second-order differential equations. In paper [5], Ma considers the following periodic boundary value problem:

{ u ( t ) + f ( t , u t ) = 0 , 0 < t < T , u 0 = φ , u ( T ) = A ,
(1.3)

and he obtained some sufficient conditions for the existence of at least one positive solution of the PBVP (1.3).

In [19], by applying the fixed point theorem of cone expression and compression, Liu investigates the existence of multiple positive solutions of the following problem:

{ u ( t ) = f ( t , u t ) , t t k , t ( 0 , T ) , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( t ) = φ ( t ) , t [ τ , 0 ] , u ( 0 ) = u ( T ) .

Motivated by the results in [5, 19, 24], the aim of this paper is to consider the existence of multiple positive solutions for the PBVP (1.1) by using some properties of the Green function and the fixed point index theorem in cones.

This paper will be divided into three sections. In Section 2, we provide some preliminaries and establish several lemmas which will be used throughout Section 3. In Section 3, we shall give the existence theorems of multiplicity positive solutions of PBVP (1.1).

2 Preliminary and lemmas

Let C τ := {φ:[τ,0]R; φ(t) is continuous everywhere except for a finite number of points t ¯ at which φ( t ¯ + ) and φ( t ¯ ) exist and φ( t ¯ )=φ( t ¯ )}, then C τ is a normed space with the norm

φ [ τ , 0 ] = sup θ [ τ , 0 ] |φ(θ)|,φ C τ .

Let J =[τ,T], C( J ) and C 1 ( J ) represent the set of a continuous and continuously differentiable on J , respectively. Moreover, for uC( J ) we define u J = sup t J |u(t)|. Furthermore, we denote:

PC( J ) = {u:u(t) is a map from J into R such that u(t) is continuous at t t k , and u( t k + ), u( t k ) exist and u( t k )=u( t k ) for k=1,2,,m; u(t)=φ(t) for t[τ,0]}.

P C 1 ( J ) = {uPC( J ):u | ( t k , t k + 1 ) C 1 ( t k , t k + 1 ), u ( t k ) and u ( t k + ) exist; and u ( t k )= u ( t k ) for k=1,2,,m}. P C + ( J )={uPC( J ):u(t)0,t[τ,T]}.

Clearly, PC( J ) is a Banach space with the norm u J = sup t J |u(t)| for u(t)PC( J ). P C 1 ( J ) is also a Banach space with the norm u 1 =max{ u J , u J }.

We need to assume the following conditions:

(A1) 1 0 T G(s,s)ds>0; φ(θ)0 for θ[τ,0], and Aϕ(0);

(A2) f(t,u)0 for t[0,T] and uP C + ( J );

(A3) I k (k=1,2,,m) are continuous and I k (u)0 for uP C + ( J ).

Lemma 2.1 (see [25])

Let E be a Banach spaces and KE be a cone in E. Let r>0 and Ω r ={xK:x<r}. Assume that S: Ω ¯ r K is a completely continuous operator such that Sxx for x Ω r .

  1. (i)

    If Sxx for x Ω r , then i(S, Ω r ,K)=1.

  2. (ii)

    If Sxx for x Ω r , then i(S, Ω r ,K)=0.

Lemma 2.2 For any y, a k PC(J,R), and η,AR. Then the problem

{ ( ρ ( t ) u ( t ) ) + y ( t ) = 0 , t J , t t k , k = 1 , 2 , , m , Δ u ( t k ) = a k , k = 1 , 2 , , m , u ( 0 ) = η , u ( T ) = A
(2.1)

has a unique solution

u(t)=η+ ( A η ) ϕ ( T ) ϕ(t)+ 0 T G(t,s)y(s)ds+ 0 < t k < T G(t, t k )ρ( t k ) a k ,
(2.2)

where

G(t,s)= 1 ϕ ( T ) { ( ϕ ( T ) ϕ ( t ) ) ϕ ( s ) , 0 s t T , ϕ ( t ) ( ϕ ( T ) ϕ ( s ) ) , 0 t s T , ϕ(t)= 0 t d s ρ ( s ) <+.
(2.3)

Proof Integrating the first equation of (2.1) over the interval [0,t] for t[0,T), we get

u(t)=u(0)+ρ(0) u (0)ϕ(t) 0 t [ ϕ ( t ) ϕ ( s ) ] y(s)ds 0 < t k < t [ ϕ ( t ) ϕ ( t k ) ] ρ( t k ) a k .
(2.4)

It follows from the boundary conditions u(0)=η, u(T)=A and (2.4) that

u (0)= 1 ρ ( 0 ) ϕ ( T ) [ ( A η ) + 0 T [ ϕ ( T ) ϕ ( s ) ] y ( s ) d s + 0 < t k < T [ ϕ ( T ) ϕ ( t k ) ] ρ ( t k ) a k ] .

Together with (2.4), we obtain (2.2). □

By the standard discussion, we have the following lemma which will be used later.

Lemma 2.3 The Green function G(t,s) defined in (2.3) has the following properties:

  1. (i)

    G(t,s)0, t,s[0,T];

  2. (ii)

    G(t,s)G(s,s)<+, t,s[0,T];

  3. (iii)

    0<σG(s,s)G(t,s), t[α,β], s[0,T], where α(0, t 1 ], β[ t m ,T), and

    0<σ:=min { ϕ ( T ) ϕ ( β ) ϕ ( T ) ϕ ( α ) , ϕ ( α ) ϕ ( β ) } <1.

Form Lemma 2.2, the problem (1.1) is equivalent to the integral equation:

(Su)(t):=u(t)= { φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( t ) + 0 T G ( t , s ) f ( s , u s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) I k ( u t k ) , t [ 0 , T ] , φ ( t ) , t [ τ , 0 ] .
(2.5)

Definition 2.4 A function uP C 1 ( J ) C 2 (J) is called a positive solution of PBVP (1.1), if it satisfies the PBVP (1.1) and u(t)0 on J , and u(t)0 on J.

Define a cone K in PC( J ) as follows:

K= { x P C ( J ) : x 0  on  J  and  x ( t ) σ x [ 0 , T ] , t [ α , β ] } .
(2.6)

Lemma 2.5 The operator S:KK is completely continuous.

Proof It is easy to show that S(K)K holds. Let B be a bounded subset in PC( J ). By virtue of the Ascoli-Arzela theorem, we only show that S(B) is bounded in PC( J ) and S(B) is equicontinuous. For any u C τ and θ[τ,0], t[0,T], we have u t (θ)=u(t+θ). Therefore, the set { u t :uB,t[0,T]} is uniformly bounded with respect to t[0,T] on C τ . Then there exist two constants L 1 >0, L 2 >0 such that

max u B , s [ 0 , T ] { | f ( s , u s ) | } < L 1 , max u B , 0 < t k < T { | I k ( u t k ) | } < L 2 ,k=1,2,,m.
(2.7)

Taking

L= max t , s [ 0 , T ] { | G ( t , s ) | } , L 0 = max t , t k [ 0 , T ] { | G ( t , t k ) | } ,k=1,2,,m.
(2.8)

It follows from (2.5), (2.7), and (2.8) that S(B) is bounded in PC( J ).

Let uB and t, t [τ,T] with t< t . There are three possibilities:

Case I. If 0t< t T, then

| ( S u ) ( t ) ( S u ) ( t ) | 0 T | G ( t , s ) G ( t , s ) | f ( s , u s ) d s + A ϕ ( 0 ) ϕ ( T ) | ϕ ( t ) ϕ ( t ) | + 0 < t k < T | G ( t , t k ) G ( t , t k ) | ρ ( t k ) I k ( u t k ) .

Case II. If τt< t 0, then we have |(Su)(t)(Su)( t )|=|φ(t)φ( t )|.

Case III. If τt<0< t T, then

| ( S u ) ( t ) ( S u ) ( t ) | | ( S u ) ( t ) ( S u ) ( 0 ) | + | ( S u ) ( 0 ) ( S u ) ( t ) | 0 T | G ( t , s ) G ( 0 , s ) | f ( s , u s ) d s + A φ ( 0 ) ϕ ( T ) | ϕ ( t ) 0 | + | φ ( 0 ) φ ( t ) | + 0 < t k < T | G ( t , t k ) G ( 0 , t k ) | ρ ( t k ) I k ( u t k ) .

Clearly, in either case, it follows from the continuity of G(t,s) and the uniform continuity of φ in [τ,0] that for any ε>0, there exists a positive constant δ, independent of t, t and u, whenever |t t |<δ, such that |(Su)(t)S(u)( t )|ε holds. Therefore, S(B) is equicontinuous. □

3 The main results

In this section, we shall consider the existence of multiple positive solutions for the periodic boundary value problems (1.1).

For convenience sake, we set

f := lim v P C + , v [ τ , 0 ] f ( t , v ) v [ τ , 0 ] , I k := lim v P C + , v [ τ , 0 ] I k ( v ) v [ τ , 0 ] ,k=1,2,,m.

For the first theorem we need the following hypotheses:

(C1) There exists a constant a 1 >0 such that for vP C + ( J ): v [ τ , 0 ] a 1 ,

f ( t , v ) M 0 max { v [ τ , 0 ] , φ [ τ , 0 ] } , t [ 0 , T ] , I k ( v ) M k max { v [ τ , 0 ] , φ [ τ , 0 ] } , k = 1 , 2 , , m ,

where M 0 and M k are two positive constants satisfying:

σ { M 0 α β G ( 1 2 , s ) d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M k } 1.
(3.1)

(C2) There exists a constant b 1 >0 satisfying:

b 1 >max { φ [ τ , 0 ] , a 1 , ( 1 D ) 1 A } ( D  be given in (3.2) )

such that

f(t,v) M 0 v [ τ , 0 ] ,t[0,T]and I k (v) M k v [ τ , 0 ] ,k=1,2,,m

for vP C + ( J ): v [ τ , 0 ] b 1 , where M 0 >0, M k >0 are constants satisfying:

D:= max t [ 0 , T ] { M 0 0 T G ( t , s ) d s + k = 1 m G ( t , t k ) ρ ( t k ) M k } <1.
(3.2)

Theorem 3.1 Assume that (C1), (C2), f =, and I k = hold. Then PBVP (1.1) has at least two positive solutions u , u with 0< u [ τ , T ] < b 1 < u [ τ , T ] .

Proof For any uK, we have u(t)σ u [ 0 , T ] , t[α,β]. It follows from the definitions of PC( J ) and u t that

u s [ τ , 0 ] = sup θ [ τ , 0 ] |u(s+θ)|u(s)σ u [ 0 , T ] .

Then for uK with u [ τ , T ] = a 1 , it follows from (2.5) and assumption (C1) that

( S u ) ( 1 2 ) = φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( 1 2 ) + 0 T G ( 1 2 , s ) f ( s , u s ) d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) I k ( u t k ) 0 T G ( 1 2 , s ) f ( s , u s ) d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) I k ( u t k ) 0 T G ( 1 2 , s ) M 0 max { u s [ τ , 0 ] , φ [ τ , 0 ] } d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) M k max { u t k [ τ , 0 ] , φ [ τ , 0 ] }
(3.3)
α β G ( 1 2 , s ) M 0 max { u s [ τ , 0 ] , φ [ τ , 0 ] } d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M k max { u t k [ τ , 0 ] , φ [ τ , 0 ] } α β G ( 1 2 , s ) M 0 max { σ u [ 0 , T ] , φ [ τ , 0 ] } d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M k max { σ u [ 0 , T ] , φ [ τ , 0 ] } σ { M 0 α β G ( 1 2 , s ) d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M k } u [ τ , T ] u [ τ , T ] = a 1 .
(3.4)

Now if we set Ω a 1 ={uK: u [ τ , T ] < a 1 }, then (3.4) shows that S u [ τ , T ] u [ τ , T ] for u Ω a 1 . Thus, Lemma 2.1 yields

i(S, Ω a 1 ,K)=0.
(3.5)

For uK with u [ τ , T ] = b 1 , from (2.5) and assumption (C2), we have

| S u ( t ) | = { | φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( t ) + 0 T G ( t , s ) f ( s , u s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) I k ( u t k ) | , t [ 0 , T ] , | φ ( t ) | , t [ τ , 0 ] { 0 T G ( t , s ) M 0 u s [ τ , 0 ] d s + 0 < t k < T G ( t , t k ) ρ ( t k ) M k u t k [ τ , 0 ] + A , φ [ τ , 0 ] { max t [ 0 , T ] { M 0 0 T G ( t , s ) u s [ τ , 0 ] d s + 0 < t k < T G ( t , t k ) ρ ( t k ) M k u t k [ τ , 0 ] } + A , φ [ τ , 0 ] { D b 1 + A , φ [ τ , 0 ] < b 1 .
(3.6)

Set Ω b 1 ={uK: u [ τ , T ] < b 1 }. Then (3.6) shows that S u [ τ , T ] < u [ τ , T ] for u Ω b 1 .

Hence, Lemma 2.1 implies that

i(S, Ω b 1 ,K)=1.
(3.7)

According to f = and I k =, choose a constant b such that

b >max { φ [ τ , 0 ] , b 1 }
(3.8)

and

f ( t , v ) M ¯ 0 v [ τ , 0 ] , t [ 0 , T ] , v P C + ( J ) , σ b v [ τ , 0 ] , I k ( v ) M ¯ k v [ τ , 0 ] , v P C + ( J ) , σ b v [ τ , 0 ] , k = 1 , 2 , , m ,

where M ¯ 0 >0, M ¯ k >0 are constants satisfying:

σ { M ¯ 0 α β G ( 1 2 , s ) d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M ¯ k } >1.
(3.9)

For uK with u [ τ , T ] = b , from (2.5), Lemma 2.5 and (3.9), by using the same method to get (3.3), we can get

( S u ) ( 1 2 ) = φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( 1 2 ) + 0 T G ( 1 2 , s ) f ( s , u s ) d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) I k ( u t k ) α β G ( 1 2 , s ) M ¯ 0 u s [ τ , 0 ] d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M ¯ k u t k [ τ , 0 ] σ { α β G ( 1 2 , s ) M ¯ 0 d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) M ¯ k } u [ τ , T ] > b .
(3.10)

Now if we set Ω b ={uK: u [ τ , T ] < b }. Then (3.10) shows that S u [ τ , T ] > u [ τ , T ] for u Ω b . Thus, an application of Lemma 2.1 again shows that

i(S, Ω b ,K)=0.
(3.11)

Since a 1 < b 1 < b , it follows from (3.5), (3.7), (3.11), and the additivity of the fixed index that

i(S, Ω b 1 Ω ¯ a 1 ,K)=1,i(S, Ω b Ω ¯ b 1 ,K)=1.

Thus, S has a fixed point u in Ω b 1 Ω ¯ a 1 , and a fixed point u in Ω b Ω ¯ b 1 . They are positive solutions of the PBVP (1.1) and

0< u [ τ , T ] < b 1 < u [ τ , T ] .

The proof is complete. □

For the second theorem we need the following hypotheses:

(C3) There exists a constant a 2 >0 satisfying

a 2 >max { φ [ τ , 0 ] , ( 1 D ) 1 A } ( D is given in (3.12) ) ,

such that for vP C + ( J ): v [ τ , 0 ] a 2 ,

f(t,v) λ 0 v [ τ , 0 ] ,t[0,T]and I k (v) λ k v [ τ , 0 ] ,k=1,2,,m,

where λ 0 , λ k are positive constants satisfying:

D := max t [ 0 , T ] { λ 0 0 T G ( t , s ) d s + k = 1 m G ( t , t k ) ρ ( t k ) λ k } <1.
(3.12)

(C4) H k : R + R + (k=1,2,,m) are continuous nonincreasing functions such that | I k (v)| H k ( v [ τ , 0 ] ) for vP C + ( J ).

(C5) There exists a constant b 2 with b 2 > a 2 such that for any vP C + ( J ):σ b 2 v [ τ , 0 ] b 2 ,

f(t,v) λ 0 max { v [ τ , 0 ] , φ [ τ , 0 ] } ,t[0,T]

and

I k (v) λ k max { v [ τ , 0 ] , φ [ τ , 0 ] } ,k=1,2,,m,

where λ 0 >0, λ k >0, and

σ { λ 0 α β G ( 1 2 , s ) d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) λ k } 1.

Theorem 3.2 Assume that (C3)-(C5), and f =0 hold. Then PBVP (1.1) has at least two positive solutions u , u with 0< u [ τ , T ] < b 2 < u [ τ , T ] .

Proof For any uK with u [ τ , T ] = a 2 . According to (2.5) and assumption (C3), we have

| S u ( t ) | = { | φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( t ) + 0 T G ( t , s ) f ( s , u s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) I k ( u t k ) | , t [ 0 , T ] , | φ ( t ) | , t [ τ , 0 ] , { max t [ 0 , T ] { λ 0 0 T G ( t , s ) u s [ τ , 0 ] d s + 0 < t k < T G ( t , t k ) ρ ( t k ) λ k u t k [ τ , 0 ] } + A , φ [ τ , 0 ] , { D a 2 + A , φ [ τ , 0 ] < a 2 .
(3.13)

Set Ω a 2 ={uK: u [ τ , T ] < a 2 }. Then (3.13) shows that S u [ τ , T ] < u [ τ , T ] for u Ω a 2 . Thus Lemma 2.1 implies

i(S, Ω a 2 ,K)=1.
(3.14)

On the other hand, it follows from (2.5) and assumption (C5) that

( S u ) ( 1 2 ) = φ ( 0 ) + ( A φ ( 0 ) ) ϕ ( T ) ϕ ( 1 2 ) + 0 T G ( 1 2 , s ) f ( s , u s ) d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) I k ( u t k ) 0 T G ( 1 2 , s ) λ 0 max { u s [ τ , 0 ] , φ [ τ , 0 ] } d s + 0 < t k < T G ( 1 2 , t k ) ρ ( t k ) λ k max { u t k [ τ , 0 ] , φ [ τ , 0 ] } α β G ( 1 2 , s ) λ 0 max { σ u [ 0 , T ] , φ [ τ , 0 ] } d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) λ k max { σ u [ 0 , T ] , φ [ τ , 0 ] } σ { λ 0 α β G ( 1 2 , s ) d s + α < t k < β G ( 1 2 , t k ) ρ ( t k ) λ k } u [ τ , T ] u [ τ , T ] = b 2 for any  u K  with  u [ τ , T ] = b 2 .
(3.15)

Set Ω b 2 ={uK: u [ τ , T ] < b 2 }. Then (3.15) implies that S u [ τ , T ] u [ τ , T ] for u Ω b 2 . Thus an application of Lemma 2.1 again shows that

i(S, Ω b 2 ,K)=0.
(3.16)

In view of assumption (C4) and f =0, there are two possibilities:

Case 1. Suppose that f is unbounded, then there exists a constant d 1 satisfying:

d 1 >max { φ [ τ , 0 ] , b 2 } ,

such that

f(t,v) ε 0 v [ τ , 0 ] for vP C + ( J ) : v [ τ , 0 ] d 1 ,t[0,T]
(3.17)

and

| I k (v)| H k ( v [ τ , 0 ] ) H k ( d 1 ),k=1,2,,m,

where ε 0 is a positive constant satisfying

max t [ 0 , T ] { ε 0 d 1 0 T G ( t , s ) d s + k = 1 m G ( t , t k ) ρ ( t k ) H k ( d 1 ) } +A d 1 .
(3.18)

If uK with u [ τ , T ] = d 1 , then it follows from (2.5), (3.17), and (3.18) that

| S u ( t ) | { max t [ 0 , T ] { ε 0 d 1 0 T G ( t , s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) H k ( d 1 ) } + A , φ [ τ , 0 ] d 1 .
(3.19)

Case 2. Suppose that f is bounded. Then there exists a constant N such that

|f(t,v)|N,t[0,T],vP C + ( J ) .

Taking

d 2 max { φ [ τ , 0 ] , max t [ 0 , T ] { N 0 T G ( t , s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) H k ( 0 ) } + A , b 2 } .

For uK and u [ τ , T ] = d 2 , from (2.5), we have

| S u ( t ) | { max t [ 0 , T ] { N 0 T G ( t , s ) d s + 0 < t k < T G ( t , t k ) ρ ( t k ) H k ( 0 ) } + A , φ [ τ , 0 ] d 2 .

Choose d=max{ d 1 , d 2 }. Hence, in either case, we always may set

Ω d = { u K : u [ τ , T ] < d } ,

such that S u [ τ , T ] u [ τ , T ] for u Ω d . Thus, Lemma 2.1 yields

i(S, Ω d ,K)=1.
(3.20)

Since a 2 < b 2 <d, it follows from (3.14), (3.16), (3.20), and the additivity of the fixed point index that

i(S, Ω b 2 Ω ¯ a 2 ,K)=1,i(S, Ω d Ω ¯ b 2 ,K)=1.

Thus, S has a fixed point u in Ω b 2 Ω ¯ a 2 , and a fixed point u in Ω d Ω ¯ b 2 . They are positive solutions of the PBVP (1.1) and

0< u [ τ , T ] < b 2 < u [ τ , T ] .

The proof is complete. □

Example 3.3 Consider the following PBVP:

{ ( e t u ( t ) ) = ( 1 + t 2 ) F ( sup θ [ 1 , 0 ] | u t ( θ ) | ) , Δ u ( 1 2 ) = I ( sup θ [ 1 , 0 ] | u ( 1 2 + θ ) | ) , u ( t ) = 1 20 ( 1 + sin t ) , t [ 1 , 0 ] , u ( 1 ) = 2 ,
(3.21)

where

F(x)= { 1 x , x [ 0 , 0.25 ) , 2 x , x [ 0.25 , 20 ) , 1 10 x 2 , x [ 20 , + ) , I(x)= { x + 0.8 , x [ 0 , 0.6 ) , 7 3 x , x [ 0.6 , 21 ) , 1 147 x 3 14 , x [ 21 , + ) .
(3.22)

Then PBVP (3.21) has at least two positive solutions u , u with 0< u [ 1 , 1 ] <10< u [ 1 , 1 ] .

Proof PBVP (3.21) can be regarded as a PBVP of the form (1.1), where

f(t,x)= ( 1 + t 2 ) F ( x ) , I 1 (x)=I ( x ) ,φ(t)= 1 20 (1+sint),
(3.23)

and ρ(t)= e t , A=2, τ=1, T=1.

First we have 1 0 1 G(s,s)ds0.8980. By choosing α=0.25, β=0.75, we get σ=0.2543<1. On the other hand, it follows from (3.21) and (3.22) that f = and I 1 = are satisfied. Finally, we show that (C1) and (C2) hold. Choose M 0 =10, M 1 =16, a 1 = 1 25 , b 1 =10, M 0 =2, M 1 = 7 3 . By calculation, we get

σ { M 0 0.25 0.75 G ( 1 2 , s ) d s + G ( 1 2 , 1 2 ) ρ ( 1 2 ) M 1 } 1.0868>1,D0.7754

and

max { φ [ τ , 0 ] , a 1 , ( 1 D ) 1 A } 8.9048< b 1 =10.

Then it is not difficult to see that the conditions (C1) and (C2) hold.

By Theorem 3.1, PBVP (3.21) has at least two positive solutions u , u with 0< u [ 1 , 1 ] <10< u [ 1 , 1 ] . □

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Acknowledgements

The authors are highly grateful for the referees’ careful reading and comments on this paper. The research is supported by Hunan Provincial Natural Science Foundation of China (Nos. 13JJ3106 and 12JJ2004); it is also supported by the National Natural Science Foundation of China (Nos. 61074067 and 11271372).

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Zhao, Y., Chen, H. & Qin, B. Periodic boundary value problems for second-order functional differential equations with impulse. Adv Differ Equ 2014, 134 (2014). https://doi.org/10.1186/1687-1847-2014-134

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Keywords

  • boundary value problem
  • impulsive functional differential equations
  • multiple solutions
  • cones